Total distance traveled is the integral of the absolute value of velocity, ∫|v(t)|dt, over a time interval. Unlike displacement, it counts every bit of motion as positive, so direction changes add to the total instead of canceling out.
Total distance traveled measures how far a particle actually moves, adding up every stretch of motion whether the particle is going forward or backward. In calculus terms, you compute it as the definite integral of the absolute value of velocity:
The absolute value is the whole point. Plain velocity is signed, so integrating v(t) directly lets backward motion cancel forward motion, and you get net displacement instead. Slapping absolute value bars on v(t) flips every negative chunk positive, so nothing cancels. Think of it as the odometer reading versus the "how far from home are you" reading. The odometer is total distance. In practice, you either let your calculator evaluate ∫|v(t)|dt directly, or you find where v(t) = 0 (the turnaround points), split the interval there, and add the absolute values of the pieces.
Total distance traveled lives in the particle motion strand of AP Calculus, where you connect position, velocity, and acceleration through derivatives and integrals (introduced with derivatives in Unit 4, then revisited with integrals in Units 6 and 8). It tests whether you really understand what a definite integral accumulates. Integrating v(t) gives net change in position; integrating |v(t)| gives total distance. That distinction is one of the most reliable point-earners (or point-losers) on the exam. In BC, the same idea extends to motion in the plane, where total distance becomes arc length, the integral of speed √((x′(t))² + (y′(t))²). Particle motion shows up on the FRQ section almost every single year, so this is a skill worth automating.
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Net Displacement (Units 6 & 8)
Displacement is ∫v(t)dt and total distance is ∫|v(t)|dt. They're equal only when the particle never changes direction. If v(t) changes sign anywhere on the interval, total distance is strictly bigger, because displacement lets backward motion cancel forward motion.
Accumulation Function (Unit 6)
Total distance is an accumulation function with |v(t)| as the rate. It's the cleanest example of the big Unit 6 idea that integrating a rate of change gives total accumulated change, here with the sign stripped off so everything accumulates positively.
Path Length / Arc Length (Unit 8, BC)
For a particle moving in the xy-plane, total distance traveled is the arc length of its path, ∫√((x′(t))² + (y′(t))²)dt. Same concept as ∫|v(t)|dt, just upgraded to 2D, since √((x′)² + (y′)²) is the particle's speed.
First Derivative (Unit 4)
Velocity is the first derivative of position, and its zeros mark where the particle turns around. Those turnaround points are exactly where you split the interval when computing total distance by hand, because they're where |v(t)| changes how it unfolds.
Particle motion is one of the most recycled FRQ setups on both AB and BC, and total distance is a standard part. The 2018 AB FRQ Q2 and 2021 FRQ Q2 each gave a velocity function for a particle on the x-axis and asked for total distance over an interval, which on the calculator-active section means setting up and evaluating ∫|v(t)|dt. On BC, the 2023 and 2024 FRQ Q2 problems moved the particle into the xy-plane, where total distance becomes the speed integral ∫√((x′(t))² + (y′(t))²)dt. To earn the points, you need to write the correct integral setup (the setup itself is usually worth a point), include the absolute value or the square root expression, and give the numeric answer to three decimal places. On no-calculator MCQs, expect simpler velocities where you find the zeros of v(t), split the interval, and add piece by piece.
Displacement asks "where did you end up relative to where you started?" while total distance asks "how far did you actually travel?" Displacement is ∫v(t)dt and can be negative or even zero if the particle returns to its start. Total distance is ∫|v(t)|dt and is always ≥ |displacement|. If an FRQ asks for total distance and you integrate v(t) without absolute value bars, you've answered the wrong question and lose the points.
Total distance traveled equals the definite integral of the absolute value of velocity, ∫|v(t)|dt, over the given time interval.
Displacement is ∫v(t)dt without absolute value, so total distance and displacement only match when the particle never changes direction.
Total distance is a scalar, meaning it's always nonnegative and direction doesn't matter, while displacement carries a sign.
Without a calculator, find where v(t) = 0, split the interval at those turnaround points, integrate each piece, and add the absolute values of the results.
On BC, total distance for motion in the plane is the arc length integral ∫√((x′(t))² + (y′(t))²)dt, which is just integrating speed.
Writing the correct integral setup, absolute value bars included, typically earns its own point on particle motion FRQs.
It's how far a particle actually moves over a time interval, computed as ∫|v(t)|dt, the integral of the absolute value of velocity. Every segment of motion counts as positive, even when the particle moves backward.
No. Displacement is ∫v(t)dt and measures net change in position, which can be negative or zero. Total distance is ∫|v(t)|dt and is always at least as large as the absolute value of displacement. They're equal only if velocity never changes sign.
Solve v(t) = 0 to find where the particle turns around, split the time interval at those points, integrate v(t) on each piece, and add the absolute values. On calculator-active sections, just evaluate ∫|v(t)|dt directly.
No. Total distance is a scalar quantity, so it's always nonnegative. If you get a negative answer, you almost certainly integrated v(t) instead of |v(t)| and actually computed displacement.
It shows up regularly in particle motion FRQs. The 2018 and 2021 exams asked for total distance of a particle on the x-axis using ∫|v(t)|dt, and the 2023 and 2024 BC exams asked for it in the xy-plane using the speed integral ∫√((x′(t))² + (y′(t))²)dt.